OFFSET
1,1
COMMENTS
abs(log(a(n)) - n - log(log(n))) < sqrt(n/log(1+n^(1/7))), but for large n, the absolute error bound of the asymptotic of log(a(n)) tends to sqrt(n/log(n)). If one tries to find the midline of the squared error, (log(a(n)) - n - log(log(n)))^2, for large n, the midline would run along pi(n) = A000720(n).
A slightly better absolute error bound could be achieved by using the imaginary part of the nontrivial zeros of the Riemann zeta function, abs(log(a(n)) - n - log(log(n))) < sqrt(((zetazero(n)-1/2)/sqrt(-1))/(3/2)). The zeta zeros also give the upper bound of the squared error, (log(a(n)) - n - log(log(n)))^2 < ((zetazero(n)-1/2)/sqrt(-1))/(3/2).
log(a(n))-sqrt(n/log(1+n^(1/7))) is a lower bound of sigma_1(n) = A000203(n). Such that, n+log(log(n))-sqrt(n/log(1+n^(1/7))) < sigma_1(n) < H(n)+exp(H(n))*log(H(n)).
a(n) gives the total number of ordered pairs (k,m) where k in set {1,2,...,n}, m in set {1,2,...,A003418(n+1)}, and k divides m. Example: For n = 3, there are 22 ordered pairs (k,m) where k is {1,2,3} and m is a multiple of k up to 12. For k = 1, every m is a multiple of 1, m is {1,2,3,4,5,6,7,8,9,10,11,12} so there are 12 pairs. For k = 2, every m is a multiple of 2, m is {2,4,6,8,10,12} so there are 6 pairs. For k = 3, every m is a multiple of 3, m is {3,6,9,12} so there are 4 pairs. So the total ordered pairs is 12 + 6 + 4 = 22 = a(3). Each ordered pair (k,m) also represents an edge in a bipartite graph. Counting all such pairs gives the total number of edges in a graph.
LINKS
J. C. Lagarias, An elementary problem equivalent to the Riemann hypothesis, Am. Math. Monthly 109 (6) (2002) 534-543. arXiv preprint, arXiv:math/0008177 [math.NT], 2000-2001.
FORMULA
a(n) = A025558(n)*(Integral_{x=0..1} Li_1(x^(1/(n+1)))/x^(1/(n+1)) dx).
Integral_{x=0..1} Li_1(x^(1/(n+1)))/x^(1/(n+1)) dx = ((n+1)/n)*H(n) = a(n)/A025558(n).
((n+1)/n)*H(n) ~ log(n) + gamma + (log(n)+gamma+1/2)/n + O(1/n^2).
log(a(n)) ~ n + log(log(n)) + O(sqrt(n/log(1+n^(1/7)))).
G.f. for ((n+1)/n)*H(n): G(x) = Li_2(x)+(1/2)*log(1-x)^2-log(1-x)/(1-x), the lim_{x->oo} G(x) = -zeta(2).
Hyperbolic l.g.f. for ((n+1)/n)*H(n): LH(x) = Li_2(x)+(1/2)*log(1-x)^2+Li_3(x)-Li_3(1-x)+Li_2(1-x)*log(1-x)+(1/2)*log(x)*log(1-x)^2+zeta(3), the Integral_{x=0..1} LH(x) dx = 2*zeta(3) = A152648.
Dirichlet g.f. for ((n+1)/n)*H(n): zeta(s+1)*(zeta(s)+zeta(s+2)).
EXAMPLE
a(n)/A025558(n) = [ 2/1, 9/4, 22/9, 125/48, 137/50, 1029/360, 2178/735, ... ]
To evaluate the integral:
For n = 1: Integral_{x=0..1} Li_1(x^(1/2))/x^(1/2) dx = Integral_{x=0..1} -log(1-x^(1/2))/x^(1/2) dx = -2 * -(Sum_{x=1..oo} 1/(x*(x+1))) = -2 * -1 = 2.
For n = 2: Integral_{x=0..1} Li_1(x^(1/3))/x^(1/3) dx = Integral_{x=0..1} -log(1-x^(1/3))/x^(1/3) dx = -3 * -(Sum_{x=1..oo} 1/(x*(x+2))) = -3 * -((1/2)*(1+1/2)) = -3 * -3/4 = 9/4.
For n = 3: Integral_{x=0..1} Li_1(x^(1/4))/x^(1/4) dx = Integral_{x=0..1} -log(1-x^(1/4))/x^(1/4) dx = -4 * -(Sum_{x=1..oo} 1/(x*(x+3))) = -4 * -((1/3)*(1+1/2+1/3)) = -4 * -11/18 = 22/9.
PROG
(PARI) a(n) = lcm(vector(n+1, i, i))*sum(i=1, n, 1/i); \\ Michel Marcus, Dec 28 2024
KEYWORD
nonn
AUTHOR
Miko Labalan, Dec 26 2024
STATUS
approved